Question

question 11 (1 point)
an electric skateboarder is travelling west on a path along the river at 5m/s. a brisk walker on the other side of the riverbank is walking east along a path at 2m/s. the two paths are 30m apart. the skateboarder is initially 500m east of the walker. after 60 seconds, how fast is the distance between the skateboarder and the walker changing?
note: we encourage you to use a scientific calculator for this question. the answer below was rounded to 4 decimal points.
-2.3387m/s
3.5877m/s
-6.5543m/s
-5.3452m/s
-10.2334m/s

Explanation:

Step1: Set up the variables

Let \(x\) be the distance the skateboarder has traveled west, \(y\) be the distance the walker has traveled east. The initial horizontal distance between them is \(D_0 = 500\) m. The vertical distance \(h=30\) m.
The distance \(s\) between them is given by \(s=\sqrt{(500 - x - y)^2+30^2}\) using the Pythagorean theorem.
We know that \(x = v_{skate}t\) (where \(v_{skate}=5\) m/s) and \(y = v_{walk}t\) (where \(v_{walk}=2\) m/s). So \(x = 5t\) and \(y = 2t\). Then \(s=\sqrt{(500-(5 + 2)t)^2+30^2}=\sqrt{(500 - 7t)^2+900}\)

Step2: Differentiate \(s\) with respect to \(t\)

Using the chain - rule, if \(s=\sqrt{u}\) where \(u=(500 - 7t)^2+900\), then \(\frac{ds}{dt}=\frac{1}{2\sqrt{u}}\cdot\frac{du}{dt}\)
First, \(\frac{du}{dt}=2(500 - 7t)(- 7)\)
When \(t = 60\) s:
\(x=5\times60 = 300\) m, \(y=2\times60=120\) m
\(u=(500-(5 + 2)\times60)^2+900=(500 - 420)^2+900=80^2+900=6400 + 900=7300\)
\(\frac{du}{dt}=2(500-7\times60)(-7)=2(500 - 420)(-7)=2\times80\times(-7)=-1120\)

Step3: Calculate \(\frac{ds}{dt}\)

\(\frac{ds}{dt}=\frac{-1120}{2\sqrt{7300}}\)
\(\sqrt{7300}\approx85.44\)
\(\frac{ds}{dt}=\frac{-1120}{2\times85.44}\approx - 6.5543\) m/s

Answer:

-6.5543m/s